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Each year the US Environmental Protection Agency (EPA) releases fuel economy data on cars manufactured in that year. Below arc summary statistics on fuel efficiency (in miles/gallon) from random samples of cars with manual and automatic transmissions manufactured in 2012. Do these data provide strong evidence of a difference between the average fuel efficiency of cars with manual and automatic transmissions in terms of their average city mileage?

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Answer:

Follows are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attachment file.

[tex]n_1 = 26 \\\\ x_1 = 16.12 \\\\s_1 = 3.58\\\\n_2 = 26\\\\ x_2 = 19.85 \\\\ s_2 = 4.51[/tex]

[tex]H_0: u_1 = u_2\\\\H_1: u_1 \neq u_2[/tex]

Using formula:

[tex]\to S^{2}_{p}=\frac{S^{2}_{1}(n_1 -1)+ S^{2}_{2}(n_2-1)}{n_1+n_2-2}[/tex]

[tex]\to Sp^2 = \frac{((26-1) \times (3.58^2) + (26-1) \times (4.51^2))}{(26+26-2)}[/tex]

          [tex]= \frac{((25) \times 12.8164 + (25) \times 20.3401 )}{(26+24)}\\\\ = \frac{(320.41 + 508.5025)}{(50)}\\\\ = \frac{(828.9125)}{(50)}\\\\=16.57825[/tex]

[tex]\to Sp = 4.0716[/tex]

Calculating the value of standard error:

[tex]\to SE = Sp \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}[/tex]

          [tex]=4.0716 \times \sqrt{\frac{1}{26}+\frac{1}{26}}\\\\=4.0716 \times \sqrt{\frac{1+1}{26}}\\\\=4.0716 \times \sqrt{\frac{2}{26}}\\\\=4.0716 \times \sqrt{\frac{1}{13}}\\\\=4.0716 \times 0.277 \\\\=1.1278332\\\\= 1.1293[/tex]

Calculating the value of test statistic:

Formula:

[tex]\to t = \frac{(x_1-x_2)}{SE}[/tex]

      [tex]= \frac{(16.12 -19.85)}{1.1293}\\\\= \frac{(-3.73)}{1.1293}\\\\=-3.303[/tex]

Calculating the value of Degree of freedom:

[tex]= n_1+n_2 -2 \\\\= 26+26 -2 \\\\= 52 -2 \\\\= 50[/tex]

Because the P-value is a 2-tailed test = 0.0018

Thus they reject H0, because of P-value < 0.05,

Yeah, certain data show clearly that the means are different.

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